week 2

MZA@UTPChemEFluidMech

Fluid Statics Fluid Statics

Week 2


OutlineOutline

• Principles of fluid statics• Pressure – depth relationships• Pressure measurement• Pressure force on surface• Buoyancy


Learning OutcomeLearning Outcome

• At the end of the chapter, you should be able to:

1. Define the term “fluid statics”2. Measure the pressure using manometers3. Determine buoyancy forces


Fluid StaticsFluid Statics

Deal with forces applied by fluidsat rest or in rigid-body motion

(Rigid body motion - fluid that is moving in such a manner that there is no relative motion between adjacent

particles)


• Example: – Water in a tank.– Water in a lake (Water actually move very slowly in the lake.

However the movement of water relative to each other is nearly zero that water is seen as “static”)


No shear stresses involved.(Assumption: fluid element moves as a rigid body,

i.e. there is no relative motion between adjacent element)

The only forces develop on the surfaces of the particles will be due to pressure


Types of pressure defined in fluid mechanics:

Absolute pressureGauge pressure

Vacuum pressure


• Absolute pressure, Pabs

• Actual pressure at a given position

• Is measured relative to absolute vacuum (absolute zero pressure)

• Gage pressure, Pgauge

• The difference between the absolute pressure and the local atmospheric pressure

• Used in most pressure-measuring device.

Pgauge

Patm

Absolute vacuum, Pabs = 0

Pabs


• Vacuum pressure, Pvac

• Pressures below Patm

• Also known as – Pgauge

Pvac


Pabs

Patm


Pgauge

Patm

Pabs

Pvac


Pabs


Pabs = Patm – PvacPabs = Patm – PvacPabs = Patm + Pgauge

Pabs = Patm + Pgauge


ExampleExample

a) A vacuum gage connected to a chamber reads 24 kPa at location where the local atmospheric pressure is 92 kPa. Determine Pabs.

b) If the absolute pressure in a tank is 20 psi at normal atmospheric pressure, determine Pgage.

c) Gage pressure readings shows a value of – 60 kPa. What does it means? And, determine Pabs.


SolutionSolution

a) Pgauge = 24 kPa, Patm = 92 kPa

Pabs = 24 + 92 = 116 kPa

b) Pgauge = Pabs - Patm

= 20-14.7= 5.3 psi

c) Pgage = – 60 kPa

It means vacuum since the value is negative.

(The same if mention as Pvac = 60 kPa)

Pabs = Pgauge+Patm

= 101.3 – 60 = 41.3 kPa


• In a stationary fluid the pressure is exerted equally in all directions and is referred to as the static pressure.

P1

P2

P4P3

Fluid

An element of fluid


• In a moving fluid, the static pressure is exerted on any plane parallel to the direction of motion.

Moving fluidP


• Pressure acting at a point in a fluid is the same at all directions.


How does the pressure in a fluid in which there are no shearing stresses vary from point to point?

To answer this question consider a small rectangular element of fluid removed from some

arbitrary position within the mass of fluid of interest


• There are two types of forces acting on this element:1. Surface forces due to the pressure,2. Body force equal to the weight of the

element.

z = z

z = 0

x

ydW

Pz=0

Pz= z


• Previous analysis apply and shows that the pressure does not depend on x or y

• Thus,as we move from point to point in a horizontal plane 1 any plane parallel to the xy plane 2 ,the pressure does not change.

• Therefore, pressure is the same horizontally.

z = z

z = 0

x

ydW

Pz=0

Pz= z


• At same elevation, pressure is the same for the same fluid at rest.


• From equilibrium of forces in z-direction,

(Pressure force)z=0 – (Pressure force)z=z – (Weight) = 0

Summation of forces in all directions = 0

(Pz=0) x y – (Pz= z) x y – x y z = 0

x y z

zPP 0zzz

z = z

z = 0

x

ydW

Pz=0

Pz= z


• Taking limit z 0

γdz

dP

Barometric equationBarometric equation

z = z

z = 0

x

ydW

Pz=0

Pz= z


• The equation is separable

• Depending on the density of the fluid, we can determine the pressure at any point of elevation.

dzdP

dP = – dz = – g dz dP = – dz = – g dz


Pressure – Depth relationshipPressure – Depth relationship

1. Incompressible Fluid• Density is constant

h gP

zzPP

zd

zd dP

1212

2

1

2

1

2

1



• For free surface Ps = Patm

• Normally take Ps = 0 (Pgage definition)• P = h = Pgage

• P increase as one go downward. • P decrease as one go upward.

hPP

hPP

zd dP

s

s

h

0

P

P

s

z = h

z = 0

Ps

P


Example 1Example 1

• Find the pressure at the bottom of a tank containing glycerin under pressure shown (glycerin = 12.34 N/m3)

50 kPa

2 m


SolutionSolution

• Assumption: glycerin is an incompressible fluid

• Pbottom = Ps + h = 50 + (12.34)(2) = 74.68 kPa


Example 2Example 2

• Determine the pressure at the bottom of an open tank containing water at atmospheric pressure.

10 m


Solution Solution

• Assumption: water is an incompressible fluid

• Pbottom = Ps + gh

= 0 + (1000)(9.81)(10) = 98.1 kPag



2. Compressible Fluid• change with pressure

• At low pressure, of most gases are approximated by ideal gas law:

P = pressure

Mw = molar mass

Ru = Universal gas constant (kJ/kmol K)

R = Gas constant (kJ/kg K)T = absolute temperature (K)

TRPM

RTP

u

w



• Through integration:

12u

w12

2

1u

w2

1

2

1u

w2

1

2

1

2

1

zzTR

gMexpPP

dz TR

gMdP

P1

dz TR

PMgdP

dz gdP


Pressure MeasurementPressure Measurement

• Objective:• Understand the principles of manometer• Learn how to calculate pressure using

manometer


In measuring pressure:In measuring pressure:

• Pressure increases as one go downward

• In calculation, has to “plus” the pressure

• Pressure decreases as one go upward

• In calculation, has to “minus” the pressure

• At the same elevation P is the same• P = 0

A B


Mercury barometerMercury barometer

• Device for measuring atmospheric pressure.

• A simple barometer consists of: – a tube more than 30 inch (760 mm) long – inserted in an open container of mercury – a closed and evacuated end pointing

upward– open end in the mercury pool – mercury extending from the container up

into the tube.

P2

P1


• The space above the liquid cannot be a true vacuum.

• It contains mercury vapor at its saturated vapor pressure

• But this is extremely small at room temperatures (e.g. 0.173 Pa at 20oC)

P2

P1


Mercury barometerMercury barometer

• Equating P at both sides:

P1 – PXY = P2

P1 – Hgh = P2

P1 – P2 = Hgh

(Since pressure exerted by mercury vapor is very small)

P2

P1

Y

X z = 0

z = 0

P1 = Hgh


A standard technique for measuring pressure involves the use of liquid columns in vertical or

inclined tubes.

Piezometer tubeU-tube manometer

Inverted U-tube manometerInclined tube manometer


Piezometer tubePiezometer tube

• One arms containing the fluid is attached to the wall of the conduit.

X

Y

1

Patm

h

Liquid A


U-tube manometerU-tube manometer

• The maximum value of P1 - P2 is limited by the height of the manometer.

Y

Fluid B

Fluid A


• To measure larger P we can choose a

manometer with greater density

• To measure smaller P with accuracy we can choose a manometer fluid which is

having a density closer to the fluid density.

Y

Fluid B

Fluid A


Inverted U-tube manometerInverted U-tube manometer

• Used for measuring pressure differences in liquids.

• The space above the liquid in the manometer is filled with air which can be admitted or expelled through the tap on the top, in order to adjust the level of the liquid in the manometer.

Y

Air

Liquid A


Inclined tube manometerInclined tube manometer

1 2

a

b

c

e

L

h2

h1

d

Fluid A

Liquid Bh3

Fluid C


ManometerManometer

Advantages• Can measure small pressure differences• Does not have to be calibrated against any standard; the

pressure difference can be calculated from first principles.

Disadvantages• Not for measuring larger pressure differences• Some liquids are unsuitable for use. Surface tension can also

cause errors due to capillary rise; this can be avoided if the diameters of the tubes are sufficiently large - preferably not less than 15 mm diameter

• Slow response; unsuitable for measuring fluctuating pressures.


ExampleExample

• A simple U-tube manometer is installed across an orifice meter. The manometer is filled with mercury (specific gravity 13.6), and the liquid above the mercury is carbon tetrachloride (specific gravity 1.6). The manometer reads 200 mm. What is the pressure difference over the manometer?


• Assuming water = 1000 kg/m3

P1 + P1X – PXX’ – PXY – PY2 = P2

P1 + A(a+h) – 0 – Bh – Aa = P2

P1 + Ah – Bh = P2

P1 – P2 = h (B – A) = (200 10-3)(1000)(9.81)(13.6 –

1.6)= 23544 Pa


ExampleExample

• The fluid shown shaded in the manometer is ethyl iodide with a specific gravity of 1.93. The heights are h1 = 100 cm and h2 = 20 cm.

a) What is the gage pressure in the tank?

b) What is the absolute pressure in the tank?

h1

h2

Tank full of air


Ptank + airgh2 – EIg(h1 + h2)= Patm

Since we’re taking gage pressure, Patm = 0 Ptank = airgh2 + EIg(h1 + h2)

= – (1.2)(9.81)(0.2) + (1.93)(1000)(9.81)(1.00 + 0.200)

= 22717.6 Pa= 22.72 kPag

Assuming the density of air is too small

Ptank = airgh2 + EIg(h1 + h2) = (1.93)(1000)(9.81)(1.00 + 0.200)= 22719.96 Pa= 22.72 kPag


Ptank, abs = 22.72 kPa + 101.3 kPa

= 124.02 kPa


ExampleExample

• The mercury manometer below indicates a differential reading of 0.30 m when the pressure in pipe A is 30-mm Hg vacuum. Determine the pressure in pipe B if the specific gravity of the oil and mercury is 0.91 and 13.6, respectively.


SolutionSolution

PA + watergh2 + Hggh3 – oilg(h1 + h2 + h3) + oilgh3 = PB

Since PA = – HggH

– HggH + watergh2 + Hggh3 – oilg(h1 + h2 + h3) + oilgh1 = PB

PB = watergh2 + Hgg (h3 – H) – oilg(h2 + h2)= (1000)(9.81)(0.15) +

(13.6)(1000)(9.81)(0.3 – 30x10-3) – (0.91)(1000)(9.81)(0.15 + 0.30)

= 33.47 kPa


ExampleExample

• A 6-in.-diameter piston is located within a cylinder which is connected to a ½ in-diameter inclined-tube manometer.The fluid in the cylinder and the manometer is oil (specific weight 59 lbf/ft3). When a weight, W, is placed on the top of the cylinder, the fluid level in the manometer tube rises from point (1) to (2). How heavy is the weight? Assume that the change in position of the piston is negligible.


SolutionSolution• Let the pressure due to piston alone (without weight) = PP

• Cross sectional area of the piston = AP

• Weight = W • Pressure due to the piston alone:

PP – oil h1 sin 30O = 0 ……………. (1)

• When the weight is put on top of the piston, new pressure due to (weight + piston), PP’

PP’ = (Initial pressure due to piston) + (Pressure due to weight)

= PP + ……………. (2)PA

W


• On the other hand, new manometer equation due to weight:

PP’ – oil(h1 + ) sin 30O = 0 ……………. (3)

(2) into (3)

PP + – oil h1 sin 30O – oil sin 30O = 0

126

PA

W

12

6


From (1) = oil sin 30O

f

2

3f

Ib 90.2

0.5ft12

6ft

12

6

4ft

Ib59W

PA

W

12

6


Mechanical and Electronic Pressure Measuring Device

Mechanical and Electronic Pressure Measuring Device

Measure of high pressures


Bourdon Pressure GageBourdon Pressure Gage

• When pressure acts on an elastic structure, the structure will deform.

• This deformation relates to the magnitude of pressure


• As the pressure within tube increases tube straighten translated into pointer motion on dial.

• The pressure indicated the difference between

that communicated by the system to the external (ambient) pressure

Pgage


Pressure transducerPressure transducer

• Converts pressure to electrical output

• Continuously monitor pressure changes with time (for rapidly changing pressure)

• Two type:– Diaphragm gage– Quartz-crystal piezometer

One side of the diaphragm is open to the external targeted pressure, PExt, and the other side is connected to a known pressure, PRef,. The pressure difference, PExt - PRef, mechanically deflects the diaphragm


Pressure measurementPressure measurement

When the static pressure in a moving fluid is to be determined, the measuring surface must be

parallel to the direction of flow so that no kinetic energy is converted into pressure energy at the

surface.


If the fluid is flowing in a circular pipe the measuring surface must be perpendicular to the radial

direction at any point.

Moving fluidD


The pressure connection, which is known as a piezometer tube, should flush to the wall of the

pipe so that the flow is not disturbed.

Moving fluidD


The pressure is then measured near the walls where the velocity is a minimum

The reading would be subject only to a small error if the surface were not quite parallel to

the direction of flow.


BuoyancyBuoyancy


BuoyancyBuoyancy

• Buoyancy arises from the fact that – fluid pressure increases with depth – the increased pressure is exerted in all

directions (Pascal's principle) – so that there is an unbalanced upward force on

the bottom of a submerged object


Buoyant force – PrincipleBuoyant force – Principle

• The buoyant force is caused by the increase of pressure in a fluid with depth.

• Consider a body with a thickness, b is submerged in a liquid of density, f.

• The hydrostatic forces:F1 = P1A acting downwardF2 = P2A acting upward

a

b

P1

P2


• Buoyant force = the difference between the two forces i.e the net upward force

FB = P2A – P1A

= f g(a+b)A – f gaA

= f gbA

= f gVb

a

b

P1

P2


The buoyant force on the solid object is equal to the weight of the fluid displaced

(Archimedes' principle)


For fully submerged or immersed body in a fluid , Archimedes’ principle is restated as:

The buoyant force on a completely submerged body is equal to the weight of

fluid displaced

FB = W = fgVb = fVbFB = W = fgVb = fVb


Buoyant force = Weight of fluid displaced

With respect to the fluid dispersed:

FB = W = bgVf = bVfFB = W = bgVf = bVf

body

fbody

body

bodyf

bodyfB

ρ

ρ gm

ρ

m gρ

gVρF


ExampleExample

• A piece of irregularly shaped metal weighs 300.0 N in air. When the metal is completely submerged in water, it weighs 232.5 N. find the volume of the metal.


SolutionSolution

• FB = W = 300 – 232.5 = 67.5 N

FB = fgV = (1000)(9.81) V = 67.5

V = 0.00688 m3


For floating bodies, Archimedes’ principle is restated as:

A floating body displaces a volume of fluid whose

weight is exactly equal to its own

The weight of the entire body must be equal to the buoyant force

(Weight of the fluid whose volume is equal to the volume of the

submerged portion of the floating body).


• For floating bodies,

f

body avg,

total

subbody,

totalbody avg,subbody,f

B

V

V

gVρ gV

WF

ρ

ρ

ρ


ExampleExample

• Determine the submerged depth of a cube of steel 0.30 m on each side floating in mercury.

Given SGsteel = 7.8

SGHg = 13.6

D

steel

mercury


SolutionSolution

• FB = W

Vbody,total = 0.3 0.3 0.3 = 0.027 m3

f gVbody,sub = body gVbody,total

(13.6)(1000)(9.81)(0.3 0.3 D) = (7.8)(1000)(9.81)(0.027)

D = 0.172 m


ExampleExample

• A solid block (SG = 0.9) floats such that 75% of its volume is in water and 25% of its volume is in liquid X, which is layered above the water. Determine the density of liquid X.


SolutionSolution

• FB = W

bgVf = fgVb

bg(Vw + VX) = wgVb,sub in w + XgVb,sub in X

bVb = wVb,sub in w + XVb,sub in X

0.9Vb = (1.0)(0.75Vb) + (SGX)(0.25Vb)

SGX = 0.60 X = 600 kg/m3


Sink or Float?Sink or Float?


Buoyancy: SubmarineBuoyancy: Submarine

week 2

Documents

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